Q: In a one-way ANOVA, prove,

$$E\left[\frac n{k-1}\sum_{i=1}^k (\bar{x}_{i.}-\bar{x}_{..})^2\right]=\sigma^2+\frac n{k-1}\sum_{i=1}^k\alpha_i^2$$

where the symbols used have their usual meanings.

Attempt 1:

Since $\mu_i=\mu+\alpha_i$, this is equivalent to saying $\bar{x}_{i.}=\bar{x}_{..}+\alpha_i$, so we have,

$$E\left[\frac n{k-1}\sum_{i=1}^k (\bar{x}_{i.}-\bar{x}_{..})^2\right]=E\left[\frac n{k-1}\sum_{i=1}^k\alpha_i^2\right]=E(U)\textrm{ (say)}$$

Somehow I need to show that $E(U)=\sigma^2+U$ but I can't seem to do this.

Attempt 2:

Using the theorem SST=SS(Tr)+SSE, we get that our problem is equivalent to proving the following:

$$E\left[\frac{SS(Tr)}{k-1}\right]=\sigma^2+\frac n{k-1}\sum_{i=1}^k\alpha_i^2$$

But then again, we know that $Z=\dfrac{SS(Tr)}{\sigma^2}$ is a chi square variate with $k-1$ degrees of freedom, so $E(Z)=k-1$ which would imply $E\left[\dfrac{SS(Tr)}{k-1}\right]=\sigma^2$ ?

I have tried a few other ideas but none of them worked and I don't have much time to type them out since I got an exam in about an hour or so.

Can someone help me with this? A small hint would work. Also, it'd be helpful if someone pointed out the flaws in my attempts.


  • 1
    $\begingroup$ On 1. "Since $\mu_i=\mu+\alpha_i$, this is equivalent to saying $\bar{x}_{i.}=\bar{x}_{..}+\alpha_i$" ... -- no, it isn't. On 2. On what basis do you assert that $Z$ is chi-square? What result are you using? $\endgroup$ – Glen_b -Reinstate Monica May 23 '17 at 9:34

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